College

Solve [tex]$4|x+5|=28$[/tex].

A. [tex]$x=12$[/tex] and [tex][tex]$x=-2$[/tex][/tex]
B. [tex]$x=12$[/tex] and [tex]$x=-12$[/tex]
C. [tex][tex]$x=-12$[/tex][/tex] and [tex]$x=-2$[/tex]
D. [tex]$x=-12$[/tex] and [tex][tex]$x=2$[/tex][/tex]

Answer :

Let's solve the equation [tex]\(4|x+5|=28\)[/tex].

1. Isolate the absolute value expression:
Divide both sides of the equation by 4 to simplify:

[tex]\[
|x+5| = \frac{28}{4} = 7
\][/tex]

2. Solve the equation without absolute value:
The equation [tex]\(|x+5| = 7\)[/tex] means that the expression inside the absolute value can be either 7 or -7. So, we set up two separate equations to solve for [tex]\(x\)[/tex]:

- Equation 1: [tex]\(x + 5 = 7\)[/tex]
Subtract 5 from both sides:
[tex]\[
x = 7 - 5 = 2
\][/tex]

- Equation 2: [tex]\(x + 5 = -7\)[/tex]
Subtract 5 from both sides:
[tex]\[
x = -7 - 5 = -12
\][/tex]

3. Conclusion:
The solutions to the equation [tex]\(4|x+5|=28\)[/tex] are [tex]\(x = 2\)[/tex] and [tex]\(x = -12\)[/tex].

Therefore, the correct answer is:
D. [tex]\(x = -12\)[/tex] and [tex]\(x = 2\)[/tex]